If the straight line x + y = 1 touches the pa | vedclass

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(C) The slope of the tangent line $x + y = 1$ is $m = -1$.For the parabola $y^2 - y + x = 0$,we differentiate with respect to $x$ to find the slope of

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$(0, 1)$

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(C) The slope of the tangent line $x + y = 1$ is $m = -1$.For the parabola $y^2 - y + x = 0$,we differentiate with respect to $x$ to find the slope of the tangent at any point $(h, k)$:$2y \frac{dy}{dx} - \frac{dy}{dx} + 1 = 0$$\frac{dy}{dx}(2y - 1) = -1$$\frac{dy}{dx} = \frac{-1}{2y - 1}$At the point of contact $(h, k)$,the slope of the tangent must equal the slope of the line:$\frac{-1}{2k - 1} = -1$$2k - 1 = 1$$2k = 2$$k = 1$Since the point $(h, k)$ lies on the line $x + y = 1$,we substitute $k = 1$:$h + 1 = 1$$h = 0$Thus,the point of contact is $(0, 1)$.

If the straight line $x + y = 1$ touches the parabola $y^2 - y + x = 0$,then the coordinates of the point of contact are

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